# Infinite Series $\sum_{n=1}^\infty\frac{H_n}{n^22^n}$

How can I prove that $$\sum_{n=1}^{\infty}\frac{H_n}{n^2 2^n}=\zeta(3)-\frac{1}{2}\log(2)\zeta(2).$$ Can anyone help me please?

• Does $H_n$ mean the n-th harmonic number? Perhaps the tag ({tag:harmonic-numbers]) is also applicable, if yes. Dec 12, 2013 at 15:58
• Yes, n-th harmonic number. Thanks for the notice.
– user95733
Dec 12, 2013 at 16:09
• There are very many similar questions linked on the right, under Related. Perhaps you can find some inspiration from the answers provided there ? Dec 12, 2013 at 16:14
• See a related problem.you need to modify it a little to solve your problem. Dec 19, 2013 at 14:14
• See this question. The Op posted a nice formula that you can use. Dec 19, 2013 at 14:22

Let's start with the product of $\;-\ln(1-x)\,$ and $\dfrac 1{1-x}$ to get the product generating function
(for $|x|<1$) : $$\tag{1}f(x):=-\frac {\ln(1-x)}{1-x}=\sum_{n=1}^\infty H_n\, x^n$$ Dividing by $x$ and integrating we get : \begin{align} \sum_{n=1}^\infty \frac{H_n}n\, x^n&=\int \frac{f(x)}xdx\\ &=-\int \frac{\ln(1-x)}{1-x}dx-\int\frac{\ln(1-x)}xdx\\ \tag{2}&=C+\frac 12\ln(1-x)^2+\operatorname{Li}_2(x)\\ \end{align} (with $C=0$ from $x=0$)
The first integral was obtained by integration by parts, the second from the integral definition of the dilogarithm or the recurrence for the polylogarihm (with $\;\operatorname{Li}_1(x)=-\ln(1-x)$) : $$\tag{3}\operatorname{Li}_{s+1}(x)=\int\frac {\operatorname{Li}_{s}(x)}x dx$$

Dividing $(2)$ by $x$ and integrating again returns (using $(3)$ again) : \begin{align} \sum_{n=1}^\infty \frac{H_n}{n^2}\, x^n&=\int \frac {\ln(1-x)^2}{2\,x}dx+\int \frac{\operatorname{Li}_2(x)}x dx\\ &=C+I(x)+\operatorname{Li}_3(x)\\ \end{align} with $I(x)$ obtained by integration by parts (since $\frac d{dx}\operatorname{Li}_2(1-x)=\dfrac {\ln(x)}{1-x}$) : \begin{align} I(x)&:=\int \frac {\ln(1-x)^2}{2\,x}dx\\ &=\left.\frac{\ln(1-x)^2\ln(x)}{2}\right|+\int \ln(1-x)\frac {\ln(x)}{1-x}dx\\ &=\left.\frac{\ln(1-x)^2\ln(x)}{2}+\ln(1-x)\operatorname{Li}_2(1-x)\right|+\int \frac{\operatorname{Li}_2(1-x)}{1-x}dx\\ &=\left.\frac{\ln(1-x)^2\ln(x)}{2}+\ln(1-x)\operatorname{Li}_2(1-x)-\operatorname{Li}_3(1-x)\right|\\ \end{align} getting the general relation : $$\tag{4}\sum_{n=1}^\infty \frac{H_n}{n^2}\, x^n=C+\frac{\ln(1-x)^2\ln(x)}{2}+\ln(1-x)\operatorname{Li}_2(1-x)+\operatorname{Li}_3(x)-\operatorname{Li}_3(1-x)$$ (with $C=\operatorname{Li}_3(1)=\zeta(3)$ here)
applied to $x=\dfrac 12$ with $\operatorname{Li}_2\left(\frac 12\right)=\dfrac{\zeta(2)-\ln(2)^2}2$ from the link returns the wished : \begin{align} \sum_{n=1}^\infty \frac{H_n}{n^2\;2^n}&=\zeta(3)-\frac{\ln(2)^3}2-\ln(2)\frac{\zeta(2)-\ln(2)^2}2\\ \tag{5}\sum_{n=1}^\infty \frac{H_n}{n^2\;2^n}&=\zeta(3)-\ln(2)\frac{\zeta(2)}2 \end{align}

• I forgot that you had used the generating function of the Harmonic Numbers here. I used its integral for the sum of $H_n^2/n^2$ recently.
– robjohn
Aug 26, 2014 at 16:49
• @robjohn: I reused these results in this answer (quite heavy machinery I'll admit...). Some appropriate references to your work are in the comments :-). Excellent continuation! Aug 26, 2014 at 21:01
• Aug 26, 2014 at 22:48
• ... corrected by Markus Scheuer here... Jul 19, 2016 at 9:48
• Is it well-known that such Harmonic sums can be succinctly expressed by the Nielsen generalized logarithm as in this post? Jun 1, 2019 at 6:43

We will make frequent use of $$\binom{n+1}{k+1}=\binom{n}{k}\frac{n+1}{k+1}\tag{1}$$ The Generalized Harmonic Numbers of the second order are defined as $$H_n^{(2)}=\sum_{k=1}^n\frac1{k^2}\tag{2}$$ The factor of $2^{-n}$ in each term reminded me of the Euler Series Transformation. Reversing the series acceleration (series deceleration?), it can be seen that we should look at \begin{align} \sum_{k=0}^n\binom{n}{k}(-1)^k\frac{H_{k+1}^{(2)}}{k+1} &=\sum_{k=0}^n\binom{n}{k}\frac{(-1)^k}{k+1}\sum_{j=0}^k\frac1{(j+1)^2}\tag{3}\\ &=\frac1{n+1}\sum_{k=0}^n(-1)^k\binom{n+1}{k+1}\sum_{j=0}^k\frac1{(j+1)^2}\tag{4}\\ &=\frac1{n+1}\sum_{j=0}^n\sum_{k=j}^n(-1)^k\binom{n+1}{k+1}\frac1{(j+1)^2}\tag{5}\\ &=\frac1{n+1}\sum_{j=0}^n(-1)^j\binom{n}{j}\frac1{(j+1)^2}\tag{6}\\ &=\frac1{(n+1)^2}\sum_{j=0}^n(-1)^j\binom{n+1}{j+1}\frac1{j+1}\tag{7}\\ &=\frac1{(n+1)^2}\sum_{j=0}^n(-1)^j\sum_{k=0}^n\binom{k}{j}\frac1{j+1}\tag{8}\\ &=\frac1{(n+1)^2}\sum_{k=0}^n\frac1{k+1}\sum_{j=0}^n(-1)^j\binom{k+1}{j+1}\tag{9}\\ &=\frac1{(n+1)^2}\sum_{k=0}^n\frac1{k+1}\tag{10}\\[3pt] &=\frac{H_{n+1}}{(n+1)^2}\tag{11} \end{align} Explanation:
$\ \;(3)$: use $(2)$
$\ \;(4)$: apply $(1)$
$\ \;(5)$: change order of summation
$\ \;(6)$: $\sum\limits_{j=k}^n(-1)^j\binom{n+1}{j+1}=(-1)^k\binom{n}{k}$
$\ \;(7)$: apply $(1)$
$\ \;(8)$: $\sum\limits_{j=0}^n\binom{j}{k}=\binom{n+1}{k+1}$
$\ \;(9)$: $\sum\limits_{j=0}^k(-1)^j\binom{k+1}{j+1}=1$
$(10)$: $H_{n+1}=\sum\limits_{k=0}^n\frac1{k+1}$

Using $(11)$, the Euler Series Transformation says that \begin{align} \sum_{k=0}^\infty(-1)^k\frac{H_{k+1}^{(2)}}{k+1} &=\sum_{n=0}^\infty2^{-n-1}\sum_{k=0}^n\binom{n}{k}(-1)^k\frac{H_{k+1}^{(2)}}{k+1}\\ &=\sum_{n=0}^\infty2^{-n-1}\frac{H_{n+1}}{(n+1)^2}\tag{12} \end{align} Therefore, with a change of indexing, we get \begin{align} \sum_{n=1}^\infty2^{-n}\frac{H_n}{n^2} &=\sum_{n=1}^\infty(-1)^{n-1}\frac{H_n^{(2)}}{n}\tag{13}\\ &=\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n^3}+\sum_{n=1}^\infty(-1)^{n-1}\frac{H_{n-1}^{(2)}}{n}\tag{14}\\ &=\frac34\zeta(3)+\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n}\sum_{k=1}^{n-1}\frac1{k^2}\tag{15}\\ &=\frac34\zeta(3)+\sum_{k=1}^\infty\sum_{n=k+1}^\infty\frac{(-1)^{n-1}}{nk^2}\tag{16}\\ &=\frac34\zeta(3)+\sum_{k=1}^\infty\sum_{n=1}^\infty\frac{(-1)^{k+n-1}}{(k+n)k^2}\tag{17}\\ &=\frac34\zeta(3)+\sum_{k=1}^\infty\sum_{n=1}^\infty(-1)^{k+n-1}\left(\frac1{k^2n}-\frac1{kn(k+n)}\right)\tag{18}\\[6pt] &=\frac34\zeta(3)-\frac12\zeta(2)\log(2)+\frac14\zeta(3)\tag{19}\\[9pt] &=\zeta(3)-\frac12\zeta(2)\log(2)\tag{20} \end{align} Explanation:
$(13)$: reindex $(12)$
$(14)$: $H_n^{(2)}=\frac1{n^3}+H_{n-1}^{(2)}$
$(15)$: apply $(2)$
$(16)$: change order of summation
$(17)$: reindex $n\mapsto k+n$
$(18)$: $\frac1{(k+n)k^2}=\frac1{k^2n}-\frac1{kn(k+n)}$
$(19)$: $\sum\limits_{k=1}^\infty\sum\limits_{n=1}^\infty\frac{(-1)^{k+n}}{kn(k+n)}=\frac14\zeta(3)$ from $(5)$ and $(7)$ of this answer
$(20)$: addition

• +1. Nice, Rob! The factor of $2^n$ in the denominator made me think of the Euler series transformation, too, but I couldn't make it go through. Also, your work starting from equation (13) is an evaluation of $A(2,1)$ from my question here, which means you've got a derivation for the last of the three sums from that question. I'd be happy if you were willing to finish off your series of answers to my question with your derivation for $A(2,1)$ here. Dec 17, 2013 at 3:52
• @MikeSpivey: I was going to do so, but dinner interrupted. Note that the justification for $(19)$ cites my answer to your question.
– robjohn
Dec 17, 2013 at 5:12
• @MikeSpivey: I have now completed the third answer to your question.
– robjohn
Dec 17, 2013 at 8:59
• Nice answer robjohn. Jul 23, 2014 at 6:35
• While $(9)$ is still true, the index should be till $n$ (in the explanation part) as per what you used in the answer. Jun 17, 2017 at 2:47

\begin{eqnarray} \sum\limits_{n=1}^\infty \frac{H_n}{n^2 2^n} = \sum\limits_{m=1}^\infty \frac{1}{m} \sum\limits_{n=m}^\infty \frac{1}{n^2 2^n} = \sum\limits_{m=1}^\infty \frac{1}{m} \int\limits_{-\infty}^0(-\xi) \frac{(1/2 \exp(\xi))^m}{1-1/2 \exp(\xi)} d\xi = \\ \int\limits_{-\infty}^0 \xi \frac{\log(1 - 1/2 \exp(\xi))}{1-1/2 \exp(\xi)} d\xi = \\ \int\limits_{1/2}^1 \left(\frac{1}{u} + \frac{1}{1-u}\right) \log(u) \left[\log(2) + \log(1-u)\right] du = \\ \zeta(3) - \frac{1}{12} \pi^2 \log(2) \end{eqnarray}

I think that all the steps are clear except for the last two ones.In the second last step I substituted for 1 - 1/2 exp(xi).The only non-trivial integrals in here are $\int \log(u)/(1-u) du$ and $\int \log(u) \log(1-u)/(1-u) du$. I compute them now. The first integral is done by expanding the denominator in a series and integrating term by term. \begin{equation} \int\limits_{1/2}^1 \frac{\log(u)}{1-u} du = \sum\limits_{p=0}^\infty \int\limits_{1/2}^1 u^p \log u du = \sum\limits_{p=0}^\infty \frac{-1+2^{-1-p}}{(p+1)^2} + \log(2) \sum\limits_{p=0}^\infty \frac{2^{-1-p}}{p+1} = -\zeta(2) + Li_2(1/2) + \log(2) Li_1(1/2) \end{equation} The second integral is done by integrating by parts and using the definition of the polylogarithmic function. \begin{equation} \int\limits_{1/2}^1 \log(u) \frac{\log(1-u)}{1-u} du = \int\limits_{1/2}^1 \log(1-u) Li_2^{'}(1-u) du = \left.\log(1-u) Li_2(1-u)\right|_{1/2}^1 + Li_3(1/2) = \log(2) Li_2(1/2) + Li_3(1/2) \end{equation} Now, the only thing that remains is to bring the results together. I am sorry but due to time constraints I am not able to do it right now. I have verified with Mathematica that all the partial results are correct.

Final Note: We can clearly see that the result is expressed though elementary functions and through polylogarithms of order not bigger than three, evaluated at 1/2.From the Wikipedia page on Polylogarithms we learn those polylogarithms at 1/2 are expressed in closed form through $\pi$, $\log(2)$ and the $\zeta$ functions. Having said that we can say that this completes the proof.

• Unfortunately, I would describe this as cheating more than slightly :( Dec 12, 2013 at 17:16

Starting with $$\frac{\ln(1-x)}{1-x}=-\displaystyle \sum_{n=1}^{\infty}H_n x^n$$

multiply both sides by $$\frac{\ln x}{x}$$ then integrate from $$x=0$$ to $$1/2$$ we get \begin{align*} I&= \int_0^{1/2}\frac{\ln x\ln(1-x)}{x(1-x)}\,dx=-\sum_{n=1}^{\infty}H_n \int_0^{1/2} x^{n-1}\ln x\ dx\\ &=-\sum_{n=1}^\infty H_n\left( \frac{\ln2}{2^n n}+\frac{1}{2^n n^2}\right)=-\frac12\ln2\zeta(2)-\sum_{n=1}^\infty \frac{H_n}{2^n n^2} \tag{1} \end{align*}

Note that we used $$\sum_{n=1}^\infty \frac{H_n}{2^nn}=\frac12\zeta(2)$$ which follows from using the generating function $$\sum_{n=1}^\infty\frac{x^n H_n}{n}=\operatorname{Li_2}(x)+\frac12 \ln^2(1-x)$$ with $$x=1/2$$ where $$\operatorname{Li_2}(1/2)=\frac12\zeta(2)-\frac12\ln^22$$

on the other hand $$\begin{equation*} I= \int_{0}^{1/2}\frac{\ln x\ln(1-x)}{x(1-x)}\ dx \overset{x\mapsto1-x}{=} \int_{1/2}^{1}\frac{\ln(1-x) \ln(x)}{x(1-x)}\ dx \end{equation*}$$ and by adding the integral to both sides, we get $$\begin{equation*} 2I= \int_{0}^{1}\frac{\ln x\ln(1-x)}{x(1-x)}\ dx =-\sum_{n=1}^\infty H_n \int_0^1 x^{n-1}\ln x \ dx =\sum_{n=1}^\infty \frac{H_n}{n^2}=2\zeta(3)\tag2 \end{equation*}$$

where the last result follows from using Euler Identity.

By combining $$(1)$$ and $$(2)$$ we obtain the closed form of $$\sum_{n=1}^\infty\frac{H_n}{2^nn^2}$$.

• Pardon me, it seems that the identity below (1) where you explain what you used is wrong since $\sum _{k=1}^{\infty }\frac{H_k}{k\:2^k}=\frac{1}{2}\zeta \left(2\right)$. Sep 8, 2020 at 1:54
• @Dennis Orton I typed ln(2) by mistake. Fixed now. Thanks Sep 8, 2020 at 2:17

\begin{align*} \operatorname{Li}_3\left(\frac{1}{2}\right)&=\int _0^{\frac{1}{2}}\frac{\operatorname{Li}_2\left(x\right)}{x}\:dx=-\operatorname{Li}_2\left(\frac{1}{2}\right)\ln \left(2\right)+\int _{\frac{1}{2}}^{1}\frac{\ln \left(x\right)\ln \left(1-x\right)}{1-x}\:dx\\[2mm] &=-\frac{1}{2}\ln \left(2\right)\zeta \left(2\right)+\frac{1}{2}\ln ^3\left(2\right)+\int _0^1\frac{\ln \left(1-x\right)\ln \left(x\right)}{x}\:dx-\int _0^{\frac{1}{2}}\frac{\ln \left(x\right)\ln \left(1-x\right)}{1-x}\:dx\\[2mm] &=-\frac{1}{2}\ln \left(2\right)\zeta \left(2\right)+\frac{1}{2}\ln ^3\left(2\right)+\sum _{k=1}^{\infty }\frac{1}{k^3}-\frac{1}{2}\ln ^3\left(2\right)-\frac{1}{2}\int _0^{\frac{1}{2}}\frac{\ln ^2\left(1-x\right)}{x}\:dx\\[2mm] &=-\frac{1}{2}\ln \left(2\right)\zeta \left(2\right)+\zeta \left(3\right)-\sum _{k=1}^{\infty }\frac{H_k}{k^2\:2^k}+\sum _{k=1}^{\infty }\frac{1}{k^3\:2^k}\\[2mm] &=-\frac{1}{2}\ln \left(2\right)\zeta \left(2\right)+\zeta \left(3\right)-\sum _{k=1}^{\infty }\frac{H_k}{k^2\:2^k}+\operatorname{Li}_3\left(\frac{1}{2}\right) \end{align*} And magically we find the value for that sum \begin{align*} \sum _{k=1}^{\infty }\frac{H_k}{k^2\:2^k}=-\frac{1}{2}\ln \left(2\right)\zeta \left(2\right)+\zeta \left(3\right) \end{align*}